0

In a textbook, the conditional probability $P(A|B)$ is well defined when $P(B)\neq 0$ and is defined as the fraction $\frac{P(A\cap B)}{P(B)}$.

I am not sure about the requirement that $P(B)\neq 0$. If the sample space $\Omega=\mathbb{R}$, $B=\mathbb{Z}$ and $A=\mathbb{N}$, then $P(B)=0$ so $P(A|B)$ is undefined. However, intuitively we understand that $P(A|B)$ should be defined as $0.5$. Is my "intuitive" conditional probability understanding correct, or should we insist that $P(A|B)$ is undefined in this case?

Zuriel
  • 5,331
  • From one hand $\mathbb{Z}$ one-to-one mapped to $\mathbb{N}$, so second is not "half" from first, and from another hand how we can understand fraction with $0$ denominator? – zkutch Jan 26 '23 at 15:30
  • Your understanding that "$P(A\mid B)$ should be defined as $0.5$" is not intuitive to me and looks likely to lead to contradictions if applied liberally. Suppose $A,B,C,D$ are full disjoint effects so any intersection is impossible. Then $P(A\mid B)+P(C\mid B)+P(D\mid B)$ would be $0.5+0.5+0.5 =1.5$ which is not a conditional probability – Henry Jan 26 '23 at 15:30
  • 1
    "If the sample space $\Omega = \Bbb R$" You lost me right there. There is no such thing as a uniform probability distribution over an unbounded set like this, and certainly not over a countably infinite set which is outright impossible. You could fix this by saying the sample space was $[-50,50]$ and we consider the uniform distribution over that, or that the sample space was $\Bbb R$ and we consider a non-uniform distribution like the normal distribution over that, etc... – JMoravitz Jan 26 '23 at 15:32
  • As to your question about conditioning on probability zero events, I'm sure this is a duplicate. Give us a moment and we'll find it for you. – JMoravitz Jan 26 '23 at 15:33
  • See also here – JMoravitz Jan 26 '23 at 15:38
  • @JMoravitz, from one book that I use, "Probability theory is concerned with situations in which the outcomes occur randomly. Generically, such situations are called experiments, and the set of all possible outcomes is the sample space corresponding to an experiment." So why cannot one say "we pick a random real number" so that the sample space is $\mathbb{R}$? – Zuriel Jan 26 '23 at 15:40
  • 1
    Because either there exists no process by which one could successfully pick a random real number in a fair uniform way, such a probability distribution cannot exist (violating several axioms)... or your description is ambiguous and the actual probability distribution hasn't been adequately stated, at which point nothing can be inferred since as an extreme example it could have been the probability distribution where we always pick the number $1$ a hundred percent of the time. in which case your $P(A\mid B)$ you were asking about would have been $1$, not $0.5$ – JMoravitz Jan 26 '23 at 15:45
  • 1
    A handwavy explanation... for continuous distributions, "the area under the curve" needs to equal $1$, for example the area under the bell-shaped curve of the normal distribution does equal $1$. If you were to try to talk about a uniform distribution over the entirety of the real number line, that doesn't work. The area under the region $[n,n+1)$ would either be zero, or would be some positive value $\epsilon$... but then the area under the entirety of the curve would have been the sum of the areas under each interval since probability is $\sigma$-additive by definition... – JMoravitz Jan 26 '23 at 15:52
  • 1
    ... and so would either have been $0+0+0+\dots \to 0$ or would have been $\epsilon+\epsilon+\dots \to \infty$, neither of which is equal to $1$, a contradiction. Note, $\sigma$-additivity only cares about adding countably many (possibly infinitely countably many) things together. There is no contradiction when talking about adding uncountably many things together as in the case of the normal distribution. – JMoravitz Jan 26 '23 at 15:53
  • @JMoravitz, thank you very much for your explanation! – Zuriel Jan 26 '23 at 15:59
  • @JMoravitz, how about "picking a random integer", which is from a countably infinite set? – Zuriel Jan 26 '23 at 16:03
  • 1
    Impossible, for the same reason. There does not exist a uniform distribution over any countably-infinite set. You could pick a random integer according to a probability distribution different than the uniform distribution however, such as according to a geometric distribution... but it will not act like what your intuition may hope for like uniform. For instance, the probability of picking an even natural number when picking from the natural numbers (including zero) according to a geometric distribution with parameter $\lambda = \frac{1}{2}$ would be $\frac{2}{3}$, not $\frac{1}{2}$ – JMoravitz Jan 26 '23 at 16:14

0 Answers0